University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 36

Answer

$[\frac{(ln2)^2}{2}]\approx0.2402$

Work Step by Step

$\int_{1}^{2} (\frac{lnx}{x}) dx=\int_{1}^{2} {lnx}\cdot \frac{dx}{x}=[\frac{(lnx)^2}{2}]_{1}^{2}=[\frac{(ln2)^2}{2}]-[\frac{(ln1)^2}{2}]=[\frac{(ln2)^2}{2}]-0=[\frac{(ln2)^2}{2}]\approx0.2402$
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